a, \(\sqrt{21}>\sqrt{20}\)

\(-\sqrt{5}>-\sqrt{6}\)

\(\Rightarrow\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

b, \(\sqrt{2}< \sqrt{3}\)

\(\sqrt{8}< \sqrt{9}=3\)

\(\Rightarrow\sqrt{2}+\sqrt{8}< \sqrt{3}+3\)

a, Vì 

\(\sqrt{21}-\sqrt{5}=2346507717\)

\(\sqrt{20}-\sqrt{6}=2022646212\)

b, Vì

\(\sqrt{2}+\sqrt{8}=4242640687\)

\(\sqrt{3}+3=4732050808\)

c, Vì

\(\sqrt{5}+\sqrt{10}=5398345638\)

\(5,3=5,3\)

P/s; Ủa tôi tưởng lớp 8 mới học về Căn thức chứ

Ta biết căn( \(\sqrt{ }\)) càng lớn thì càng chia ra số nhỏ

=> a >

b<

c>

\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}}\)

\(A^2=8+2\sqrt{10+2\sqrt{5}+8-2\sqrt{10+2\sqrt{5}}+}2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)

\(A^2=16+2\left[64-4\left(10+2\sqrt{5}\right)\right]\)

\(A^2=16+128-8\left(10+2\sqrt{5}\right)\)

\(A^2=144-80-16\sqrt{5}\)

\(A^2=64-16\sqrt{5}\)

\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2.\sqrt{10+2\sqrt{5}}+2\sqrt{64-4\left(10+2\sqrt{5}\right)}\)

\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}+2^2}\)

\(=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}=16+2\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)

\(=2+2.\sqrt{2}.\sqrt{10}+10\)

\(=\left(\sqrt{2}+\sqrt{10}\right)^2\)

=> \(A=\sqrt{2}+\sqrt{10}\)

Câu hỏi của Nguyen Phuc Duy - Toán lớp 9 - Học toán với OnlineMath

Bạn tham khảo link này!

\(10+\sqrt{60}+\sqrt{24}+\sqrt{40}=10+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}\)

\(=\left(5+2\sqrt{15}+3\right)+2+2\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\)

\(=\left(\sqrt{5}+\sqrt{3}\right)^2+2\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)+2\)

\(=\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)^2\)

\(\Rightarrow\sqrt{10+\sqrt{60}+\sqrt{24}+\sqrt{40}}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)

Dùng hẳng đẳng thức 3 số:

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
$VT=\sqrt{5+3+2+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}}=\sqrt{(\sqrt5+\sqrt3+\sqrt2)^2}=VP(đpcm)$

Biến đổi vế trái ta có :

 \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)

= \(\sqrt{2}\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)\)

Đặt A  = \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

A^2 = \(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)

    =  8 + \(2\sqrt{16-\left(10-2\sqrt{5}\right)}\)

     = \(8+2\sqrt{16-10+2\sqrt{5}}\)

     = \(8+2\sqrt{6+2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)

=> A = \(\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

=> \(\sqrt{2}A=\sqrt{2}\left(\sqrt{5}+1\right)=\sqrt{10}+\sqrt{2}=VP\) ( ĐPCM) 

haha        

bn thang tran lm sai bước đưa ra hdt :v đúng là phải 16 - ( 10 + 2can5 )

= 16 - 10 - 2can5

\(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5-2\sqrt{2}.\sqrt{5}+2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\)

\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right)\sqrt{5}-\left(3\sqrt{\frac{1}{10}}+10\right)=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\frac{3\sqrt{10}}{10}-10\)

\(=-3\sqrt{10}+10-\frac{3\sqrt{10}}{10}-10=-3\sqrt{10}-\frac{3\sqrt{10}}{10}=-3\sqrt{10}\left(1+\frac{1}{10}\right)=\frac{-33\sqrt{10}}{10}=-3,3\sqrt{10}\)